## 1.8.3 Central Forces in Three Dimensions

Consider the motion of a particle in a potential field $V(r)$ in three dimensions using spherical coordinates, $r, \theta, \varphi$, where $\theta$ is the colatitude and $\varphi$ is the longitude. The kinetic energy is:

$$T(t; r, \theta, \varphi; \dot{r}, \dot{\theta}, \dot{\varphi} = \frac{1}{2} m \left(\dot{r}^2 + r^2\dot{\theta}^2 + r^2(\sin\theta)^2\dot{\varphi}^2 \right)$$
(defn T3-spherical [m]
(fn [[_, [r, theta, _], [rdot, thetadot, phidot]]]
(* 1/2 m
(+ (square rdot)
(square (* r (sin theta) phidot))))))

;; Lagrangian = T-V
(defn L3-central [m Vr]
(defn Vs [[_, [r, _, _], _]]
(Vr r))
(- (T3-spherical m) Vs))

#'user/L3-central

;; The generalized forces can be computed by taking the partial derivative of the Lagrangian w.r.t the coordinates
(rendertex ( ((partial 1) (L3-central 'm (literal-function 'V)))
(up 't
(up 'r 'theta 'phi)

;; Here \varphi is a "cyclic coordinate" as it does not appear in the Lagrangian explicitly
;; and hence does not have a force associated with it

\begin{bmatrix}\displaystyle{m\,{\dot {\phi}}^{2}\,r\,{\sin}^{2}\left(\theta\right) + m\,r\,{\dot {\theta}}^{2} - DV\left(r\right)} \cr \cr \displaystyle{m\,{\dot {\phi}}^{2}\,{r}^{2}\,\sin\left(\theta\right)\,\cos\left(\theta\right)} \cr \cr \displaystyle{0}\end{bmatrix}
;; Compute the momenta by taking partial derivative w.r.t generalized velocities
(rendertex ( ((partial 2) (L3-central 'm (literal-function 'V)))
(up 't
(up 'r 'theta 'phi)

\begin{bmatrix}\displaystyle{m\,\dot r} \cr \cr \displaystyle{m\,{r}^{2}\,\dot {\theta}} \cr \cr \displaystyle{m\,\dot {\phi}\,{r}^{2}\,{\sin}^{2}\left(\theta\right)}\end{bmatrix}

The momentum conjugate to $\varphi$ is conserved. We can show that this is actually the $z$ component of the angular momentum vector $r \times (m\vec{v})$, for position $\vec{r}$ and linear momentum $m\vec{v}$ by writing the $z$ component of the angular momentum in spherical coordinates:

;; z component of ang momentum
(defn ang-mom-z [m]
(fn [[_, xyz, v]]
(ref (cross-product xyz (* m v)) 2)))

;; Coordinate conversion
(defn s->r2 [[_, [r, theta, phi]]]
(let [x (* r (sin theta) (cos phi))
y (* r (sin theta) (sin phi))
z (* r (cos theta))]
(up x y z)))

(rendermd
((compose (ang-mom-z 'm) (F->C s->r))
(up 't
(up 'r 'theta 'phi)

;; results in m phidot r² sin²(θ) which is equal to the momentum conjugate of phi

$$m\,\dot {\phi}\,{r}^{2}\,{\sin}^{2}\left(\theta\right)$$

Since the choice of $z$ axis arbitrary based on the coordinate system, if one component is conserved, then all components are conserved. Therefore, angular momentum is conserved. We can choose the $z$ axis such that all the angular momentum is in the $z$ component.

So for a general position vector $\vec{x}$, since $\vec{x}\cdot L = \vec{x}\cdot(m\vec{x}\times\vec{v}) = \vec{v}\cdot(\vec{x}\times\vec{x}) = 0$ ( from the scalar triple product), the motion is confined to the plane perpendicular to the angular momentum i.e., colatitude $\theta = \pi/2$ and $\dot{\theta} = 0$. This motion was discussed before in Section 1.6

Computing the energy from the Lagrangian, we can see that it does equal $T+V$

(rendermd
((Lagrangian->energy (L3-central 'm (literal-function 'V)))
(up 't
(up 'r 'theta 'phi)

$$\frac{1}{2}\,m\,{\dot {\phi}}^{2}\,{r}^{2}\,{\sin}^{2}\left(\theta\right) + \frac{1}{2}\,m\,{r}^{2}\,{\dot {\theta}}^{2} + \frac{1}{2}\,m\,{\dot r}^{2} + V\left(r\right)$$